# Is the wavefunction of particles inside a gas spread or localized?

For an individual free particle that starts localized, the wave function packet spreads over time, so the particle becomes less localized. Suppose now that we have a gas of those particles inside a box, and we allow them to collide (using some potential): will the wave function of each particle still spread indefinitely, or do collisions act as a source of decoherence and the wave function relocalizes again? I have heard both arguments by different colleagues, even in textbooks. Has anybody made a computer simulation that shows what would be the best picture?

• I asked a similar question here but got no conclusive answers, so I think it must be pretty subtle. I hope you get a good canonical answer here. – knzhou Jul 3 '16 at 18:33
• It would be interesting if you cited some of the different arguments in textbooks that you've seen. – Rococo Jul 3 '16 at 20:18
• One good answer to this question would acknowledge the identical nature of particles, use second quantization, and compute the two point density correlation for a (ideal) gas. – DanielSank Jul 3 '16 at 21:58
• This is a highly non-trivial problem. I would suggest you started by solving the scattering problem for identical particles. I found a discussion of the problem in Landau, Quantum Mechanics, par. 137. – valerio Jul 4 '16 at 8:51
• Related: en.wikipedia.org/wiki/Anderson_localization which is a system that does have localization. The question is if it applies in the context, at which point I have to say the question is too broad. It certainly still has my +1 though and could be confined e.g. by using @DanielSank's comment. – Wolpertinger Jul 4 '16 at 18:36

Preliminaries: How do we define 'localized?'

For a single particle, or for multiple non-entangled particles, it is easy to tell from the expressions for the wavefunctions whether they are localized or delocalized. For example, you might say that if the wavefunction is falling off exponentially or faster for large $x$, that is with a form like $\psi(x)\sim e^{-x/\xi}$ with some characteristic length scale $\xi$, then it is localized, while something like a plane wave (which can be considered to be in the $\xi \rightarrow \infty$ limit) is delocalized.

For interacting particles, the many-body quantum state will generically evolve to something that is entangled between the particle. Then there is no longer a wavefunction for an individual particle, and the question of localization is no longer so straightforward. For example, is the two-particle state $\psi(x_1,x_2)\sim e^{-(x_1-x_2)^2}$ localized or delocalized?

A standard way to generalize this idea of localized/extended to many-body systems is by using the concept of entanglement entropy (1), and asking if a particular region is entangled with another distant region of the system. For a one-dimensional system, the entanglement entropy is:

$S(\rho_A)=-\text{Tr}[\rho_A \log \rho_A]$, with $\rho_A$ the reduced density matrix for that system:

$$\rho_A(x_1,x_2,\ldots x_N,x'_1,x'_2,\ldots x'_N)=\int_{|x_i|>x_0} \int_{|x'_i|>x_0} ~\mathrm dx_1 \ldots \mathrm dx_N~\mathrm dx'_1 \ldots \mathrm dx'_N ~\psi(x_1,\ldots, x_N) \psi^*(x'_1,\ldots, x'_N)$$

Here we are looking at a region from $-x_0$ to $x_0$. If $S$ is exponentially small, then the system is localized, and if it is not it is extended. Notice that we've moved from talking about particles to talking about regions. This is more natural when thinking about localization, but for a uniform density of particles at a particular moment in time localization of one implies the other.

The sense of "localized" that we now have is that for a localized system, a measurement at one point does not perturb the quantum state at a faraway point. Using this standard, if you carry out the above calculations on a state like the above two-particle state, or a single particle plane wave state, you will find that they have non-zero entanglement entropy and are extended. However, a state like $\psi(x_1,x_2)\sim e^{-(x_1/\xi)^2}e^{-((x_2-2x_0)/\xi)^2}$ would be localized, as long as $x_0 \gg \xi$.

Eigenstate thermalization

Okay, with these ideas in place I can now state the answer simply: for a quantum system of particles in a box that interact with a hard-shell repulsion, in a highly excited state and dilute limit, and at equilibrium, the entanglement entropy is proportional to the volume of the system.

What this means, roughly speaking, is that every point in the box is equally entangled with every other point. In this sense, the system is extended. Measuring the quantum state at one point will also affect the quantum state at every other point.

The proof of this is basically due to Srednecki, in a foundational paper of quantum thermalization which I encourage you to take a look at (2). For the above system, Srednecki shows that the eigenstates of the system give particle behavior that agrees with Maxwell-Boltzmann statistical mechanics, and furthermore that systems that start far from equilibrium (such as the case you mention where everything starts out localized) will also evolve to an equilibrium state that obeys these predictions. Furthermore, subsequent work has emphasized that any system that has this self-equilibrating property, known as eigenstate thermalization, will also necessarily show volume scaling of entanglement entropy (see, for example, (3)).

Decoherence

All of what I've said so far has been about the pure quantum state, but people often talk about this kind of system in terms of decoherence. What's the connection?

Well, decoherence happens when the system of interest is entangled with many other inaccessible systems- and that's clearly what happens here (4, 5). Since any part of the system is entangled with every other one, for a system of even moderate size it would be practically impossible to observe the coherence between different parts. This means that the system will be functionally indistinguishable from a system with no coherence, or just a classical statistical ensemble. This is the miracle of entanglement- if you have enough of it, things get simpler instead of more complicated. That's why measurements, which invariably produce some complicated entangled state between the system and apparatus, can nonetheless result in gaining knowledge.

Conclusion

There are two valid ways one can describe the state of the box of colliding particles after a long time:

1. It is a complex many-body entangled state in which each part is equally entangled with every other part, but in such as way as to reproduce standard statistical results (such as the Maxwell-Boltzmann distribution) for a single-particle measurement.
2. Because of the high amount of entanglement, for all practical purposes the particles may also be treated as decohered classical particles, in which case they of course have a well-defined position and momentum.

Neither of these claims is incorrect, and each might be useful in the right context.

It really depends on the boundary conditions. For boundary conditions like a 3D box with reflecting walls, the initial quantum state $\Psi$ will stay a quantum state with the unique wave function depending on variables of each particle: $$\Psi({\bf{r}}_1,...,{\bf{r}}_n, t).$$ If the boundary conditions are such that allow exchange with the environment, then the density matrix approach may become more appropriate.

In both cases the positions of particles are statistically predicted, but in the latter case there may not be interference phenomenon (or it will be less pronounced).

Consider your case for a double slit experiment without and with particle position measurements, measurements before the screen.

To solve your problem exactly, you would have to solve the Schrödinger equation

$$i \frac{\partial}{\partial t} \Psi (\vec r_1 \dots \vec r_N,t)= H \ \Psi(\vec r_1 \dots \vec r_N,t)$$

where $\Psi (\vec r_1 \dots \vec r_N,t)$ is the wave function of the $N$ particles and

$$H=\sum_i^N \frac{p_i^2}{2 m} + \sum_{i<j}^N u_{ij}+V_{\text{ext}}$$

where $u_{ij}$ is some pair potential and $V_{\text{ext}}$ is the box potential. Of course, you also need an initial condition

$$\Psi (\vec r_1 \dots \vec r_N,0) = \tilde \Psi (\vec r_1 \dots \vec r_N)$$

The first thing that you must notice is that it is inappropriate to talk about "the wave function of each particle", because you have to consider the total wave function of the $N$ particles ($\Psi$). If the particles are indistinguishable, this function must posses some symmetries, depending on what kind of particles you are considering (bosons or fermions).

It is really difficult to solve such a problem analytically or numerically. A good starting point to get a qualitative idea would be to solve the corresponding equations for two particles and see what happens.

This has been done numerically by the authors of this article using a gaussian and square potentials, with distinguishable and indistinguishable particles. in the latter case, symmetrized (bosonic) and antisymmetrized (fermionic) wave functions were considered:

$$\Psi'(x_1,x_2) = \frac 1 {\sqrt{2}} [\Psi(x_1,x_2)\pm \Psi(x_2,x_1)]$$

The wave function at $t=0$ is assumed to be the product of two gaussian wave packets:

$$\Psi(x_1,x_2, t=0) = g(x_1,x_1^0,k_1,\sigma) \ g(x_2,x_2^0,k_2,\sigma)$$

Where

$$g(x,x^0,k) = e^{i k x} e^{-\frac{(x-x^0)^2}{4 \sigma^2}}$$

and $k_2 = - k_1$.

In the following plot you can for example see the result of the collision of two distinguishable particles with equal mass $m$ interacting via a square potential (the curves are the probability densities obtained by the wave functions):

The numbers in the top left corners indicate the time (in units of $100 \times$ time step) and the edges of the frames correspond to the walls of the box.

You can see that there is indeed a "spreading" of the wave packets (to be more precise, thier modulus squared, that is, the probability densities) after the collision (cfr. 1 and 120).

Quoting from the article:

In Fig. 5 we show nine frames from the movie of a repulsive m–m collision in which the mean kinetic energy equals one quarter of the barrier height. The initial packets are seen to slow down as they approach each other, with their mutual repulsion narrowing and raising the packets up until the time (50) when they begin to bounce back. The wavepackets at still later times are seen to retain their shape, with a progressive broadening until they collide with the walls and break up.

In the context of solid-state physics, a closely related question has been an area of active research in the past few years. Most interacting systems do indeed thermalize (and thus delocalize) over long time scales. However, certain systems whose disorder is much stronger than their interactions experience "Many-Body Localization," in which the individual particles remain "stuck" indefinitely. This has many macroscopic consequences, like lack of electrical conduction (because the electrons aren't free to move) and entanglement entropy that only scales as an area law rather than a volume law for every eigenstate (not just the ground state). There are far too many papers on this topic to list, but the original paper that started it all is Basko, Aleiner, and Altschuler (2006).

A dense gas with relatively strong interactions (by gas standards) might roughly be thought of as a very strongly disordered, weakly interacting solid, so I suspect that just as in the solid-state case, your thought experiment might lead to either a delocalized or a localized state, depending on the details of the density of the gas and the strength of its interactions.

• I'm not convinced that a dense gas of a single species with strong interactions can be thought of as equivalent to static disorder. In one case the system has translational invariance, and in the other it does not. However, an interesting intermediate case is that in which there are two species of particles, one much heavier than the other, in which one might imagine that the heavy particles are "almost" static and look like disorder to the lighter ones. This is discussed a bit, for example, here: arxiv.org/abs/1606.05650 . – Rococo Jul 4 '16 at 18:19
• (but +1 for mentioning MBL as the other possibility for an interacting system) – Rococo Jul 4 '16 at 18:30

Quantum effects appear if the concentration of particles satisfies, $$\frac{N}{V} \ge n_q$$ where $N$ is the number of particles, $V$ is the volume, and $n_q$ is the quantum concentration, for which the interparticle distance is equal to the thermal de Broglie wavelength, so that the wavefunctions of the particles are barely overlapping. As the quantum concentration depends on temperature, high temperatures will put most systems in the classical limit unless they have a very high density e.g. a White dwarf or the very early Universe.

The quantum nature of the particle manifests itself in that bosons obey Bose–Einstein statistics and fermions obey the Fermi–Dirac statistics. Both Fermi–Dirac and Bose–Einstein statistics are well approximated by the classical Maxwell–Boltzmann statistics at high temperature and low concentration where the quantum effects are negligible.

I suppose you mean that the gas is contained in a magic box. Otherwise the walls become part of the system, exchanging momentum/energy with the 'particles'. I have no answer for you; I don't know. What I do know is that none of the particle-particle collisions can be characterized other than by using a probability distribution. Common sense demands that assuming you know a particles approximate position and momentum at time t=0, and assuming the distribution of other particles in the box is random and they are all at thermal equilibrium with the walls, the position or momentum will be more and more evenly distributed over the volume (phase space) at times t>0. So, you tell me, are you using the particle as the origin of the coordinates of said wave equation, or is the frame of reference fixed by the box? As far as I can see, your question isn't useful and lacks any predictive value. It seems to me your question is about the state of the wave equation in between observations. Which (maybe because I'm a fan of the Copenhagen Interpretation) seems profoundly misguided. You probably can get the particle localized, once (t=0) but then what? How do you observe it a 2nd time? Clearly that 2nd observation will not be at an equivalently localized place. (barring some trap).

The behaviour of the molecules in your gedanken experiment can be approached by using decoherence. But I do not believe you can get a definitive answer until somebody makes a full scale simulation (or until some expert's answer can make a formal proof of what really happens, but I am not skilled to do that). The decoherence effects can be argued heuristically, but taking that approach the answer I found is ambiguous.

On one hand, a single molecule is not isolated, it interacts with a macroscopic environment (gas, electromagnetic radiation, walls). In most cases this environment will act to reduce the entangled state of as single molecule and make it to apparently collapse into a preferred basis. The prefered basis is usually the position basis, but this is not always the case and depends on the details of the set up.

If the preferred basis is the position basis, then the wavefunction of individual molecules will not spread for too long, its interaction with the rest of the environment in which it is situated will make the probabilities to "collapse" to a smaller packet (as mentioned by P. Shor in the comment, it cannot be a pure eigenstate of the position operator). But if the preferred basis is different, for instance the momentum basis, then the end result will be a largely delocalized wave packet, This second option seems to be more consistent with the fact that once reaching thermodynamic equilibrium, the gas molecules satisfy the Maxwell Boltzmann statistics.

• The wavefunction can never collapse entirely to the position basis, as this would mean the momentum would be unbounded. – Peter Shor Jul 3 '16 at 20:25
• If you are suggesting that molecular or container interactions would result in wave function collapse, I don't think that's true. – anon01 Jul 4 '16 at 18:48
• @ConfusinglyCuriousTheThird no, that is not what I mean, I will re write my answer to try to make it more clear – Wolphram jonny Jul 4 '16 at 18:51